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Hello,

I usually see the following theorem without seeing any reference or explanation

Theorem: Let X be a blowup of $P^2$ at generic r points. If $r\leq 8$ then there are only finitely many (-1) curves on $X$ , while if $r\geq 9$ there are infinitely many of them.

This is a little mysterious to me. I imagine the proof would be finding a family of curves in $P^2$ containing these points which are centers of the blowup, so that the multiplicities of those curves at these points are large enough to make the strict transform of these curves (-1) curves. However, I don't know what family to be chosen. Can someone give a proof or a reference containing a proof of this fact?

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2  
Does Hartshorne Chapter V, Exercise 4.15(e) count? –  Will Sawin Sep 30 '12 at 5:35

4 Answers 4

up vote 8 down vote accepted

It is a result of Nagata but it is easy to explain (at least to se the difference between $r\le 8$ and $r\ge 9$):

The Picard group of the blow-up is generated by $L$, the pull-back of a general line, and $E_1,...,E_r$ the exceptional curves. Any curve disctinct from the $E_i$'s is linearly equivalent to $C=dL-\sum a_i E_i$ for some integers $d,a_1,\dots,a_r$ with $d>0$, which is the degree of the image of the curve in $\mathbb{P}^2$. The integer $a_i=E_i\cdot C$ is the multiplicity of this curve at $a_i$ and is then non-negative.

If $C$ is a $(-1)$-curve, we have $C^2=-1$ and $C\cdot C+C\cdot K_S=-2$ (adjunction formula) so

$\sum a_i=3d-1$

$\sum (a_i)^2=d^2+1$

By Cauchy-Schwartz we have $(\sum a_i)^2\le r\sum (a_i^2)$, which yields $9d^2-6d+1\le rd^2+r$. For $r\le 8$, we find thus only finitely many possiblities for $d$ ($d\le 6$), hence only finitely many $(-1)$-curves and for $r\ge 9$ we get infinitely many solutions.

It remains to see that the set of infinitely many numeric solutions gives infinitely many curves in the case $r\ge 9$. By a dimension count, we find that $|C|$ is not empty for each numeric solution. The fact that the points are in general position "should" imply that $|C|$ counts in fact an irreducible element. (This last part is the only one where something has to be checked; it seems natural but is not so easy, see the Nagata and Harbourne-Hirschowitz conjecture).

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First consider the case when the $9$ points is the base locus of a pencil of cubics.

Take the family of cubics through these $9$ points. On the blow up they form a basepoint-free linear system that defines an elliptic fibration. The exceptional curves give sections, and you can use the group structure on the elliptic curves to translate these curves. Any translate of any exceptional curve is another $(-1)$-curve and clearly there are infinitely many of them.

Now consider a family of $\mathbb P^2$'s and move the above special $9$ points into general position and consider the blown-up family of surfaces. According to Kodaira exceptional subvarieties are stable, so all the nearby members of the (blown-up) family has to have infinitely many (−1)-curves.

This means that the statement is true in an open neighbourhood of the $9$ points in $\mathbb P^2$.

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4  
@Sandor : Unless I'm mistaken, through nine general points, there will be only one cubic (for instance, because the space of cubics is 9-dimensional). The construction you give applies for particular sets of 9 points : the base loci of pencils of cubics. –  Olivier Benoist Sep 30 '12 at 6:06
    
Olivier, you're right, I didn't read the question carefully –  Sándor Kovács Sep 30 '12 at 7:03
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...but I think the above deformation argument fixes it. –  Sándor Kovács Sep 30 '12 at 14:37
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Jim: I'm not sure I understand your comment, in particular, the statement that there are only 3 points. The exceptional curves give 9 points on the generic fibre of the fibration. Choosing one of them as the origin, the others generate a subgroup of the group of k(P^1)-rational points on that curve. Unless the points are in very special position (in fact, unless they are the Hesse configuration), the group they generate will be infinite –  Artie Prendergast-Smith Oct 1 '12 at 13:26
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@Artie: I meant "three-torsion" not "three torsion". They are the nine 3-torsion points and they are closed under multiplication. –  Jim Bryan Oct 1 '12 at 21:07

This result is exactly Theorem 4a of Nagata's paper "On rational surfaces, II". Nagata's terminology and notations are a little bit old-fashioned, but it is still very legible !

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You already have plenty of references, but let me give you another one.

The standard quadratic transformation is the birational map $\sigma: \mathbf{P}^2 \dashrightarrow \mathbf{P}^2$ defined by

$$ \sigma: [x_0, x_1, x_2] \rightarrow [x_1x_2, x_0x_2, x_0x_1].$$

This maps fails to be defined at the three coordinate points; in the classical terminology, one says it is based at those points. Blowing up the plane at the three coordinate points, $\sigma$ then lifts to an automorphism of the blowup; moreover, conjugating $\sigma$ by elements of $PGL(3)$, one gets an automorphism of the blowup of $\mathbf{P}^2$ at any triple of (non-collinear) points.

The idea is then to compose maps of this kind (based at different triples of points) to produce $(-1)$ curves on the blowup of $\mathbf{P}^2$ whose degree downstairs is arbitrarily large. That is, start with any $(-1)$-curve $C$ you like, for example the proper transform of the line through two of your points. Then apply a sequence of Cremona transformations such that the successive images of $C$ are curves whose degree gets larger at every step.

But how do we know such a sequence exists? This can be translated into a problem about root systems and Weyl groups. The key point turns out to be that for the blowup of $r \leq 8$ points the root system has finite Weyl group, whereas for $r=9$ the group is $W\left(\tilde{E}_8\right)$, which is infinite.

For details about the last paragraph, the source I know is the paper "Weyl Groups and Cremona Transformations", by Dolgachev.

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2  
This is a nice description, but one has to take care: in general there is no automorphism on the blow-up of $r\ge 9$ pts. The application of the quadratic maps is good, but it could move the points in a bad way. In fact, the application of an element of the Weyl group to a $(-1)$-curve gives an element $C$ of the Picard group, which satisfies $C^2=-1$ and $CK=-1$. In most case, it should yield another $(-1)$-curve, but it could also happen that the divisor is not irreducible. Anyway, for most elements it works, by the result of Nagata. –  Jérémy Blanc Sep 30 '12 at 18:06
    
Hi Jérémy, yes, I glossed over some subtleties in the answer. One should choose the points to be in Cremona general position, meaning that they are in linear general position, and remain so after any sequence of Cremona transformations. Also, as you say, the Cremona transformations are not automorphisms of a fixed blowup. –  Artie Prendergast-Smith Sep 30 '12 at 19:19

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